2 votes
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Laplacian on the sphere of radius $r$

Your definition for $\text{grad}(f)$ is fine, however, the expression you gave for $\Delta f$ only works if you fix a point $p\in M$ and consider a 'geodesic frame at $p$', i.e an orthonormal frame $\{...
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2 votes
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Does the Laplace equation have solutions that are not separable?

It is certainly not true that every solution to the Laplace equation is separable; the sum of two different separable solutions generally won't be, for example. But the Laplace equation is linear, so ...
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2 votes

Why is a diffeomorphism an isometry if and only if it commutes with the Laplacian?

Let $\phi: M \to N$ and suppose $(M,g)$ and $(N,h)$ are Riemannian manifolds. Suppose that $\phi^* \circ \Delta_h = \Delta_g \circ \phi^*$. Then for each pair of smooth functions $u,v: N \to {\mathbb ...
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1 vote
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Asymptotic bound of the number of eigenvalues in an interval $[n - \epsilon, n + \epsilon]$ with the Weyl's law

What you have done looks good so far. Recall, we say that $f=O(g)$ if $\vert f(n) \vert \leqslant C \vert g(n) \vert$ for some $C>0$ provided that $n$ is sufficiently large. Hence, if $f_1=O(\sqrt{...
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1 vote

Help with a PDE

The first thing I would do is let $u(x,y)= p(x,y)- B$ so that $A\left(\frac{\partial^2 p}{\partial x^2}+ \frac{\partial^2 p}{\partial y^2}\right)= P- B$ becomes $A\left(\frac{\partial^2 u}{\partial x^...
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1 vote

Help with a PDE

Set $q(x, y) = p(x, y)+B$. Then, $$ Δ q = \frac{\partial^2 q(x, y)}{\partial x^2}+\frac{\partial^2 q(x, y)}{\partial y^2} = \frac{\partial^2 p(x, y)}{\partial x^2}+\frac{\partial^2 p(x, y)}{\partial y^...
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  • 1,300
1 vote
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Find that of $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ equals $= \frac{\partial^2 f}{\partial u^2} + ...$

The keyword here is "chain of rule". Consider a function $z=f(u,v)$ such that $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$ there exists and $u=g(x,y)$ and $v=h(x,y)$ ...
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1 vote

How can I solve Laplace Tranformation of $1/s^{5/2}$?

Using TravorLZH's idea we have for $x>-1$, \begin{align*} \mathcal{L}\{t^x\}&=\int_0^\infty t^xe^{-st}dt,\quad\text{doing }u=st\\ &=\int_0^\infty \left(\frac{u}{s}\right)^xe^{-u}\frac{du}{s}...
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  • 1,490

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